Answers
Answer:
Weight = 8.162 Newton.
Explanation:
Given the following data;
Mass = 2.2 kg
Acceleration due to gravity = 3.71 N/kg
To find the weight of the textbook;
Weight = mass * acceleration due to gravity
Weight = 2.2 * 3.71
Weight = 8.162 N
Therefore, the weight of the science textbook in mars is 8.162 Newton.
Related Questions
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answers
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
A 2.0 kg breadbox on a fric-
tionless incline of angle u 40 is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k 120 N/m, as shown in
Fig. 8-43. The box is released from rest when the spring is unstretched. Assume that the pulley is massless and frictionless. (a) What is the speed of the box when it has moved 10 cm down the in- cline? (b) How far down the incline from its point of release does the box slide before momentarily stopping, and what are the (c) magnitude and (d) direction (up or down the incline) of the box’s acceleration at the instant the box momentarily stops?
Answers
An electron is travelling in the positive x direction. A uniform electric field is in the negative y direction. If a uniform magnetic field with the appropriate magnitude and direction also exists in the region, the total force on the electron will be zero. The appropriate direction for the magnetic field is:Group of answer choicesthe negative y directioninto the pageout of the pagethe negative x directionthe positive y direction
Answers
Answer:
into the page
Explanation:
Since the uniform electric field is in the negative y direction so its is -E and the electron is travelling in the positive x direction, it experiences an electric force F = -e × -E = + eE, so the electric force is in the positive y direction. Now since the net force on the electron is zero in the region of the magnetic field, it follows that the direction of the magnetic force is opposite to that of the electric force. Since the electric force is in the positive y direction, the magnetic force is in the negative y direction.
By the right hand rule, since the magnetic force is in the negative y direction and the electron moves in the positive x direction, it follows that the magnetic field is in the positive z direction, into the page.
if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³
Answers
Mass=100x0.02=2
Please help me!
8. Give an example of a poor blackbody radiator and explain why it is not a good blackbody radiator.
9. Does a blackbody radiator emit light waves? Explain.
Answers
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Answer:
A black body radiator is an idealized body that absorbs all incoming electromagnetic radiation (thus the name of "black body").
A black body radiator is an object that has a lot of thermal energy, and it irradiates its thermal energy in the form of black body radiation (thermal radiation emitted by a black body).
a) Then, we could go to the trivial case of a mirror, a mirror is a poor blackbody radiator because a mirror reflects most of the incoming electromagnetic radiation, thus, a mirror is a really bad approximation for a black body, then a mirror is a poor black body radiator.
b) Any electromagnetic wave is a light wave (there exists "light" that we can not see). A black body radiator irradiates energy, and this radiation is in the form of electromagnetic waves, which are in essence, light waves.
Explanation:
The moment of inertia of the club head is a design consideration for a driver in golf. A larger moment of inertia about the vertical axis parallel to the club face provides more resistance to twisting of the club face for off-center hits. The mass of one club head is 200 g and its moment of inertia is 5000 g cm2 . What is the radius of gyration of this club head
Answers
Answer:
Explanation:
Moment of inertia I = M k² , where M is mass and k is radius of gyration .
Putting the given values in the equation
5000 = 200 x k²
k² = 25
k = 5 cm .
Radius of gyration is 5 cm .
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Answers
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]
On substituting the given values, we get
⇒ [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]
⇒ [tex]=6666.67 \ N[/tex]
A carnival ride starts at rest and is accelerated from an initial angle of zero to a final angle of 6.3 rad by a rad counterclockwise angular acceleration of 2.0 s2 What is the angular velocity at 6.3 rad?
Answers
The final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Final angular velocity of the carnival ride
The final angular velocity of the carnival ride is determined by applying third kinematic equation as shown below;
ωf = ωi + 2αθ
where;
ωf is the final angular velocity of the carnival ride = ?ωi is the initial angular velocity of the carnival ride = 0α is the angular acceleration = 2.0 rad/s²θ is the angular displacement of the carnival ride = 6.3 radωf = 0 + 2(2.0) x 6.3
ωf = 25.2 rad/s
Thus, the final angular velocity of the carnival ride at a displacement of 6.3 rad is 25.2 rad/s.
Learn more about angular velocity here: https://brainly.com/question/6860269
Answer: 5.0 rad/s
Explanation: Because that’s what khan said so try it out.
A 85-W lamp is connected to 100 V. What is the resistance of the lamp?
Answers
R=V2/P =100x100/85 =117.64 Ohum
The kinetic energy and the potential energy of the cannonball is constantly ________ as it travels through the air.
A. Changing
B. Increasing
C. Constant
D. Decreasing
Answers
Answer:
C. Constant
Explanation:
The total energy of the cannonball remains constant as it travels through the air.
Answer:
Explanation:
hi my name is Ava
What is the frequency of a monochromatic light used in a diffraction experiment that has a wavelength of 6.38 ✕ 10e-07 m?
Answers
Answer:
[tex]f=4.70\times 10^{14}\ Hz[/tex]
Explanation:
Given that,
The wavelength of light, [tex]\lambda=6.38\times 10^{-7}\ m[/tex]
We need to find the frequency of the light. We know that,
[tex]c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{6.38\times 10^{-7}}\\\\f=4.70\times 10^{14}\ Hz[/tex]
So, the required frequency of light is equal to [tex]4.70\times 10^{14}\ Hz[/tex].
Balance the equation by choosing the correct coefficient numbers in the drop down menus.
[Select]
SO2 +
[Select]
VH₂ →
[Select]
S +
[ Select]
H20
It is suggested you write this on scratch paper and balance it before choosing your answers :)
Answers
Answer:
SO₂ + 2H₂ —> S + 2H₂O
The coefficients are: 1, 2, 1, 2
Explanation:
SO₂ + H₂ —> S + H₂O
The above equation can be balance as follow:
SO₂ + H₂ —> S + H₂O
There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by writing 2 before H₂O as shown below:
SO₂ + H₂ —> S + 2H₂O
There are 2 atoms of H on the left side and 4 atoms the right side. It can be balance by writing 2 before H₂ as shown below:
SO₂ + 2H₂ —> S + 2H₂O
Now, the equation is balanced.
The coefficients are: 1, 2, 1, 2
Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
Answers
Answer:
0.03 A
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
From ohm's law, the voltage, current and resistor are related by the following formula:
Voltage = current × resistor
V = IR
With the above formula, we can obtain the current in the circuit as follow:
Voltage (V) = 12 V
Resistor (R) = 470 Ω
Current (I) =?
V = IR
12 = I × 470
Divide both side by 470
I = 12 / 470
I = 0.03 A
Thus, the current in the circuit is 0.03 A
Answer:
0.03 A
Explanation:
Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit
0.03 A
Review please help.
Answers
Answer:
1 and 3
Explanation:
because they are going up from 0
g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest at the top of an incline plane. The height of top of the incline plane is 1.2 m. Find translational speed of each cylinder upon reaching the bottom and determine which cylinder has the greatest translational speed upon reaching the bottom. Moment of inertia of hollow cylinder about its axis passing through the center is mr2 and for solid cylinder mr2/2
Answers
Answer:
a. i. 3.43 m/s ii. 2.8 m/s
b. The thin-walled cylinder
Explanation:
a. Find translational speed of each cylinder upon reaching the bottom
The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy
So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.
i. translational speed of thin-walled cylinder upon reaching the bottom
So, For the thin-walled cylinder, I = mr², we find its translational velocity, v
So, mgh = 1/2mv² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²)(v/r)²
mgh = 1/2mv² + 1/2mv²
mgh = mv²
v² = gh
v = √gh
v = √(9.8 m/s² × 1.2 m)
v = √(11.76 m²/s²)
v = 3.43 m/s
ii. translational speed of solid cylinder upon reaching the bottom
So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'
So, mgh = 1/2mv'² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²
mgh = 1/2mv'² + mv'²
mgh = 3mv'²/2
v'² = 2gh/3
v' = √(2gh/3)
v' = √(2 × 9.8 m/s² × 1.2 m/3)
v' = √(23.52 m²/s²/3)
v' = √(7.84 m²/s²)
v' = 2.8 m/s
b. Determine which cylinder has the greatest translational speed upon reaching the bottom.
Since v = 3.43 m/s > v'= 2.8 m/s,
the thin-walled cylinder has the greatest translational speed upon reaching the bottom.
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answers
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:
Answers
A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,
• the net horizontal force on the car is
∑ F = N cos(45°) = m a = m v ² / R
• the net vertical force on the car is
∑ F = N sin(45°) - m g = 0
where
• N = magnitude of the normal force
• m = mass of the car
• a = v ² / R = centripetal acceleration of the car
• v = tangential speed of the car
• R = 100 m = radius of curvature
• g = 9.8 m/s² = acceleration due to gravity
From the net vertical force equation, we get
N = m g / sin(45°)
and substituting this into the net horizontal force equation and solving for v gives
(m g / sin(45°)) cos(45°) = m v ² / R
v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s
We have that A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
From the question we are told
a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:
Generally the equation for the Velocity is mathematically given as
[tex]V=\sqrt{rgtan\theta}[/tex]
Therefore
[tex]V=\sqrt{rgtan\theta}\\\\V=\sqrt{100*9.8*tan45}\\\\V=32m/s[/tex]
Therefore
A highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of
V=32m/s
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A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answers
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
What kind of energy is in a moving skateboard
Answers
Answer:
I guess it is kinetic energy
Answer:
kinetic energy because my dog told me
Example of the center of the gravity
Answers
Answer:
The example of the center of the gravity is the middle of a seesaw
Explanation:
I hope this will help you and plz mark me brainlist
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answers
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answers
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Explain how conduction, convection, and radiation occur involving a campfire
Answers
Answer:
https://wtamu.edu/~cbaird/sq/2015/02/26/when-i-sit-by-a-campfire-how-does-its-hot-air-heat-me/#:~:text=When%20you%20sit%20by%20a,It%20comes%20from%20thermal%20radiation.&text=Since%20air%20is%20a%20good,of%20pockets%20of%20heated%20fluid.
Here's a link to help you hope it helps have a good day
Anatomy and Phys PLEASE HELP
you’ve learned about the structures that make up the reproductive system. Imagine that you are so interested in reproductive health that you are exploring careers related to this health field.
For this activity you are going to research careers related to the promotion or treatment of reproductive health and choose one career to focus on. You are going to identify the key factors of the educational path that you would need to take to work in this career, the annual salary that you would earn, the schedule that you may find yourself working, physical requirements for the job, and how long it will take you to achieve this goal.
You will produce a list of 10 or more strengths and weaknesses that will identify why you are qualified for this role.
You will then prepare a list of at least five questions that you would like to ask someone already working in the career. Your questions ought to be relevant to the career you have chosen and should be related to the research that you have conducted.
Answers
Answer:
Get to work.
Explanation:
This is not Brainly material. This is a project that you need to get started on and do your best on. It always looks worst before you get started on it. Just sink your teeth in and you'll be finishing it up before you know it.
Answer:
careers related to the promotion or treatment of reproductive health and choose one career to focus on. You are going to identify the key factors of the educational path that you would need to take to work in this career, the annual salary that you would earn, the schedule that you may find yourself working, physical requirements for the job, and how long it will take you to achieve this goal.
Explanation:
The moon does not stay at the same distance from the earth.why?
Answers
Answer:
The moon does not stay at the same distance of the earth because the ortbit of the moon is slightly elliptical. If earth is not tilted at an angle of 66.5°, there will be no change in the season and the earth will have equal length of days and night.
Explanation:
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Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at rest and while in motion. In the first experiment, the object is pushed with a constant known force along a horizontal surface. There is negligible friction between the surface and the object. A motion sensor is used to measure the speed of the object as it is pushed. In a second experiment, the object is tied to a string and pulled upward with a constant known force, and a motion sensor is used to measure the speed of the object as it is pulled upward. The student uses the data collected from the motion sensor to determine the mass of the object in both experiments.
Required:
What classifies the type of mass that was determined in each experiment?
Answers
Answer:
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
Explanation:
It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.
In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
As per the given problem, the student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass. In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.Thus, we can conclude that the in the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
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A woman shouts at a boy who is underwater what happens to the speed of the sound wave as it moves from the air into the water
Answers
Answer:
B. it increases
Explanation:
As shown in the table provided, the speed of sound in water (1493 m/s) is greater than the speed of sound in air (346 m/s).
Answer:
B is the correct answer.
Explanation:
A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
seconds, what was his acceleration?
Answers
Answer:
a = 52s²
Explanation:
How to find acceleration
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Solve
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
When a 20 kg explosive detonates and sends a 5 kilogram piece traveling to the right at 105 m/s
what is the speed and direction of the other 15 kilogram piece of the explosive!
Answers
Answer:
speed: 35m/s
direction: left
Explanation:
Assuming the right side is the positive direction:
before explosion:
P = mv = 0
after explosion:
P' = 15P + 5P
(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')
P' = 0.75mv1' + 0.25mv2'
P' = (15kg)v' + (5kg)(105m/s)
P' = 525kg/m/s + (15kg)v1'
P = P'
525kg/m/s + (15kg)v1' = 0
(15kg)v1' = -525kg/m/s
v1' = -35m/s
speed = |-35| = 35m/s
direction is to the left since the right side is the positive direction.
I need help with this review question. I’ll give extra points.
Answers
Answer:
1.9m/s²
Explanation:
Use the equation v=u+at, where v is the final speed, u is the initial speed, a is the acceleration and t is the time.
v=u+at
15.3=0+a(8)
a=15.3/8
a= 1.9125 m/s²
